The question is taking from Arithmetic of Elliptic Curves by Silverman, Q5.12 on page 154.
I've managed to show the supersingular case when $p=3 \mod 4$, which was done more generally for elliptic curves of the form $y^2=x^3+ax$.
I'm now trying to work the case when $p=1 \mod 4$. Following the proof on page 149, I have managed to obtain $A_p=((p-1)/2) choose (p-1)/4)$ like Example 4.5 on page 152. I've conveniently replaced $q$ in the book with $p$ as in the question.
What I don't get is that page 149 then says that $\#E(\mathbb{F}_p)=1-A_p$, which firstly, will give me a negative value since $A_p$ is most likely greater than 1 and secondly, nowhere near the result I'm intending to get.
I also worked out some examples just to see if I'm making a careless mistake but that doesn't seem to be the case.
$A_{13}=20$ and $\#E(\mathbb{F}_{13})=20$
$A_{17}=70$ but it's impossible for $\#E(\mathbb{F}_{17})$ to be that large by the trivial upper bound.
What am I misunderstanding here?
The fact is that $\# E(\mathbb{F}_q) \equiv 1 - A_q \bmod p$, but this is not an equality, just a congruence. Notice that after writing $\# E(\mathbb{F}_q) = 1-A_q$, Silverman does mention in p. 149:
In your case:
$p=13$, we have $A_{13}=20$, and $\# E(\mathbb{F}_{13})=20$, and $$1-A_{13}\equiv 1-20\equiv -19\equiv 7 \equiv 20 \equiv \# E(\mathbb{F}_{13})\bmod 13,$$ and
$p=17$, we have $A_{17}=70$, and $\# E(\mathbb{F}_{17})=16$, and $$1-A_{17}\equiv 1-70\equiv -69\equiv 16 \equiv \# E(\mathbb{F}_{13})\bmod 17,$$ so everything works out as it should... in terms of congruences modulo $p$.